{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW8_F09

# HW8_F09 - Stat 219 Stochastic Processes Homework Set 8 Fall...

This preview shows pages 1–3. Sign up to view the full content.

Stat 219 - Stochastic Processes Homework Set 8, Fall 2009, Due: November 18th 1. Exercise 5.2.5. Show that E ( τ β,α ) = αβ by applying Doob’s optional stopping theorem for the uniformly integrable stopped martingale W 2 t τ β,α - t τ β,α . ANS: We have seen en-route to (5.2.2) that τ β,α τ α < almost surely. Considering the martingale X t = W 2 t - t of continuous sample path we have further assumed in the statement of the exercise that X t τ β,α is U.I. Thus, Doob’s optional stopping theorem (Theorem 4.3.17) applies here, leading to the identity E ( W 2 τ β,α - τ β,α ) = E ( W 2 0 - 0) = 0. That is, E τ β,α = E W 2 τ β,α = α 2 P ( W τ β,α = α ) + β 2 P ( W τ β,α = - β ) = α 2 β α + β + β 2 α α + β = αβ. 2. Exercise 5.2.7 (a) Suppose a random vector Y is independent of random vectors Z 1 , Z 2 and Z 1 has the same law as Z 2 . Show that then Y + Z 1 has the same law as Y + Z 2 . ANS: A simple computation using characteristic functions and independence yields, E exp( i ( θ , Y + Z 1 )) = E exp( i ( θ , Y ) + i ( θ , Z 1 )) = E exp( i ( θ , Y )) E exp( i ( θ , Z 1 )) (independence) = E exp( i ( θ , Y )) E exp( i ( θ , Z 2 )) ( Z 1 L = Z 2 ) = E exp( i ( θ , Y ) + i ( θ , Z 2 )) (independence) = E exp( i ( θ , Y + Z 2 )) . We conclude that the characteristic functions of Y + Z 1 and Y + Z 2 agree and hence the two random vectors have the same distribution. (b) Suppose { Y t } and { Z t } are independent stochastic processes and { Z t } has the same distribution as {- Z t } . Explain why X t = Y t - Z t has the same distribution as W t = Y t + Z t . ANS: This immediately follows by applying part (a) to the finite dimensional distributions of { Y t } , { Z t } and {- Z t } . (c) Let W t be a Brownian motion and τ 0 non-random. Upon verifying that part (b) applies for Y t = W t τ and Z t = W t - W t τ deduce that X τ t = W t τ - ( W t - W t τ ) is a Brownian motion. ANS: To apply (b) we need to show that { Y t } and { Z t } are independent stochastic processes. Let {G t } denote the canonical filtration of { W t } and note that τ is (trivially) a stopping time for {G t } . Note that Y t is measurable with respect to G τ for all t 0. Furthermore, Z t = 0 when t τ and Z t = W t - W τ for t τ . Hence it follows from Proposition 5.2.3 that Z t is independent of {G τ } 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
and hence of Y t . Therefore by part (b), { X τ t } = { Y t - Z t } L = { Y t + Z t } = { W t } , so that { X τ t } has the same finite dimensional distributions as a Brownian motion. Since its sample path t X τ t is continuous, the desired result follows.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

HW8_F09 - Stat 219 Stochastic Processes Homework Set 8 Fall...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online